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A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
Tớ làm lần lượt nhé.
Ta có:\(\frac{3}{x-1}=\frac{4}{y-2}=\frac{5}{z-3}\)
\(\Rightarrow\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta được:
\(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}=\frac{\left(x-1\right)+\left(y-2\right)+\left(z-3\right)}{3+4+5}=\frac{\left(x+y+z\right)-\left(1+2+3\right)}{12}=\frac{18-6}{12}=1\)
\(\Rightarrow\frac{x-1}{3}=1\Rightarrow x=4\)
\(\frac{y-2}{4}=1\Rightarrow y=6\)
\(\frac{z-3}{5}=1\Rightarrow z=3\)
\(\frac{x-y}{2}=\frac{x+y}{12}=\frac{xy}{200}=\frac{x-y+x+y}{2+12}=\frac{2x}{14}=\frac{x}{7}=k\)
\(\Rightarrow x=7k\left(1\right);x+y=12k\left(2\right);xy=200k\left(3\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow y=12k-7k=5k\)
\(\Rightarrow xy=5k\cdot7k=35k^2\left(4\right)\)
Từ \(\left(3\right);\left(4\right)\Rightarrow200k=35k^2\Leftrightarrow200=35k\Leftrightarrow k=\frac{200}{35}\)
\(\Rightarrow x=7\cdot\frac{200}{35}=40\)
\(y=5\cdot\frac{200}{35}=\frac{1000}{35}\)
P/S:số khá xấu.sợ sai.nhưng cách làm là như vậy.
Đặt \(\frac{x}{2}=\frac{y}{3}=k\)\(\left(k\ne0\right)\)
=> x=2k , y =3k
x.y=54 => 2k.3k=54 => 6k^2=54
=> k=\(+-3\)
=> (x,y)=(6,9) = (-6,-9)
bài 1:
a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow2x=46\)
\(\Leftrightarrow x=23\)
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=7\cdot9\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-8\end{array}\right.\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Leftrightarrow\left(x+4\right)^2=5\cdot20\)
\(\Leftrightarrow\left(x+4\right)^2=100\)
\(\Leftrightarrow x+4=10\)
\(\Leftrightarrow x=6\)
a) \(\frac{x-3}{x+5}=\frac{5}{7}\) điều kiện x khác -5
<=> 7(x-3)=5(x+5)
<=> 7x-5x=25+21
<=> x=23
vậy x=23
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)điều kiện x khác 1
<=> 63=x2-1<=> x=\(\pm\)8
vậy x={-8;8}
c) \(\frac{x+4}{20}=\frac{5}{x+4}\) điều kiện x khác -4
<=> (x+4)2=25
<=> \(\left[\begin{array}{nghiempt}x+4=5\\x+4=-5\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=1\\x=-9\end{array}\right.\)
vậy x ={1;-9}
\(\frac{x-1}{4}=\frac{2x+1}{5}\)
\(\Rightarrow5\left(x-1\right)=4\left(2x+1\right)\)
\(\Rightarrow5x-5=8x+4\)
\(\Rightarrow5x-8x=4+5\)
\(\Rightarrow-3x=9\)
\(\Rightarrow x=-3\)
vậy_
\(\frac{x+2}{x-1}=\frac{x-3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\)
\(\Rightarrow x^2+x+2x+2=x^2-3x-x+3\)
\(\Rightarrow x^2+x+2x-x^2+3x+x=3-2\)
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
vậy_